In KMn04 molecule, the oxidation states of the elements potassium (K), manganese (Mn) and oxygen (O) are respectively

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Q: 85 (NDA-II/2012)
In KMn04 molecule, the oxidation states of the elements potassium (K), manganese (Mn) and oxygen (O) are respectively

question_subject: 

Science

question_exam: 

NDA-II

stats: 

0,3,4,2,3,0,2

keywords: 

{'elements potassium': [0, 0, 0, 1], 'oxidation states': [0, 0, 0, 1], 'manganese': [2, 0, 0, 1], 'molecule': [1, 0, 0, 9], 'oxygen': [3, 0, 1, 0], 'mn': [0, 0, 1, 2]}

In the KMnO4 molecule, the oxidation state of potassium (K) is +1. This is because potassium is an alkali metal and it typically loses one electron to achieve a stable electronic configuration of noble gas.

The oxidation state of manganese (Mn) in KMnO4 is +7. This can be determined by considering that the oxygen atoms in KMnO4 each have an oxidation state of -2, and the overall molecule is neutral. Using this information, we can calculate the oxidation state of manganese as +7 to balance out the -2 oxidation state of each oxygen atom.

The oxidation state of oxygen (O) in KMnO4 is -2. This is a common oxidation state for oxygen atoms in most compounds, as oxygen is highly electronegative and tends to gain two electrons to achieve a stable electronic configuration.

So, the correct oxidation states for potassium, manganese, and oxygen in the KMnO4 molecule are +1, +7, and -2 respectively.

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